# Continuous Functions

• Apr 28th 2010, 03:13 AM
acevipa
Continuous Functions
This question seems fairly logical if you think about it as the two functions must intersect. How would you go about proving it?

$Suppose\ that\ f\ and\ g\ are\ continuous$ $functions\ on\ the\ closed\ interval\ [a,b]$. $Show\ that\ there\ exists\ a\ point\ c \in [a,b]$ $such\ that\ f(c)=g(c)\ if\ f(a) \leq g(a)\ and\ f(b) \geq g(b)$
• Apr 28th 2010, 03:20 AM
HallsofIvy
Quote:

Originally Posted by acevipa
This question seems fairly logical if you think about it as the two functions must intersect. How would you go about proving it?

$Suppose\ that\ f\ and\ g\ are\ continuous$ $functions\ on\ the\ closed\ interval\ [a,b]$. $Show\ that\ there\ exists\ a\ point\ c \in [a,b]$ $such\ that\ f(c)=g(c)\ if\ f(a) \leq g(a)\ and\ f(b) \geq g(b)$

Look at h(x)= f(x)- g(x).
• Apr 28th 2010, 03:32 AM
acevipa
Quote:

Originally Posted by HallsofIvy
Look at h(x)= f(x)- g(x).

Sorry, I still don't quite understand.
• Apr 28th 2010, 04:09 AM
Defunkt
What is the sign of h(a)? What is the sign of h(b)?
• Apr 28th 2010, 04:12 AM
acevipa
Quote:

Originally Posted by Defunkt
What is the sign of h(a)? What is the sign of h(b)?

Ok I see, so h(c)=0 as intercepts the x-axis. But how would you go about proving it still?
• Apr 28th 2010, 05:46 AM
acevipa
Would you say $h(a) \leq 0\ and h(b) \geq 0$

$Therefore,\ there\ exists\ a\ c \in (a,b)\ such\ that\ h(a)=0$
• Apr 28th 2010, 07:50 AM
Defunkt
Yep, that is exactly the intermediate value theorem: Intermediate value theorem - Wikipedia, the free encyclopedia
Note, though, that since you have a weak inequality, it is possible that $c \in [a, b]$ ie. $c=a$ or $c=b$.